mathematic 求解微分方程DSolve[{x'[t]== r*x[t]+b,x[0] == x0},x[t],t]为什么出现错误:DSolve::deqn:Equation or list of equations expected instead of True in the first argument {-b+(x^\[Prime])[t]==r x[t],True}.当我把x换成y时,即DSo
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![mathematic 求解微分方程DSolve[{x'[t]== r*x[t]+b,x[0] == x0},x[t],t]为什么出现错误:DSolve::deqn:Equation or list of equations expected instead of True in the first argument {-b+(x^\[Prime])[t]==r x[t],True}.当我把x换成y时,即DSo](/uploads/image/z/14443652-20-2.jpg?t=mathematic+%E6%B1%82%E8%A7%A3%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8BDSolve%5B%7Bx%27%5Bt%5D%3D%3D+r%2Ax%5Bt%5D%EF%BC%8Bb%2Cx%5B0%5D+%3D%3D+x0%7D%2Cx%5Bt%5D%2Ct%5D%E4%B8%BA%E4%BB%80%E4%B9%88%E5%87%BA%E7%8E%B0%E9%94%99%E8%AF%AF%EF%BC%9ADSolve%3A%3Adeqn%3AEquation+or+list+of+equations+expected+instead+of+True+in+the+first+argument+%7B-b%2B%28x%5E%5C%5BPrime%5D%29%5Bt%5D%3D%3Dr+x%5Bt%5D%2CTrue%7D.%E5%BD%93%E6%88%91%E6%8A%8Ax%E6%8D%A2%E6%88%90y%E6%97%B6%2C%E5%8D%B3DSo)
mathematic 求解微分方程DSolve[{x'[t]== r*x[t]+b,x[0] == x0},x[t],t]为什么出现错误:DSolve::deqn:Equation or list of equations expected instead of True in the first argument {-b+(x^\[Prime])[t]==r x[t],True}.当我把x换成y时,即DSo
mathematic 求解微分方程DSolve[{x'[t]== r*x[t]+b,x[0] == x0},x[t],t]
为什么出现错误:DSolve::deqn:Equation or list of equations expected instead of True in the first argument {-b+(x^\[Prime])[t]==r x[t],True}.
当我把x换成y时,即DSolve[{y'[t] == r*y[t] + b,y[0] == y0},y[t],t]时却能成功求解,这是为什么?
mathematica求解微分方程中,x只能作自变量,y只能作因变量吗?例如求解x(t),则出现错误。
mathematic 求解微分方程DSolve[{x'[t]== r*x[t]+b,x[0] == x0},x[t],t]为什么出现错误:DSolve::deqn:Equation or list of equations expected instead of True in the first argument {-b+(x^\[Prime])[t]==r x[t],True}.当我把x换成y时,即DSo
我把你的代码复制到mathematic里面:
DSolve[{x'[t]== r*x[t]+b,x[0] == x0},x[t],t]
运行报错,但是如果这么写
DSolve[{x'[t] == r*x[t] + b,x[0] == x0},x[t],t]
就能求出解来
看出上面两句有什么不同了吗?
我靠!你的上面那行的加号是中文输入法里的加号,这不管在哪种程序里都是错的啦.